Question

$P$ and $Q$ are any two points lying on the sides $DC$ and $AD$ respectively of a parallelogram $ABCD$. Show that $area\left(\Delta APB\right)=area\left(\Delta BQC\right)$

Answer 1

$△APB$ and ||gm $ABCD$ are on the same base $AB$ and between same parallel $AB$ and $DC$.
Using the property,

$area\left(\Delta APB\right)=\frac{1}{2}area$(||gm $ABCD$) ...................................(1)

Similarly,

$area\left(\Delta BQC\right)=\frac{1}{2}area$(||gm $ABCD$) ..................................(2)

From (1) and (2)

we have Area $\left(\Delta APB\right)=$ Area $\left(\Delta BQC\right)$